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Problem:

After adding a $\pu{150 mL}$ $\pu{6.7\times10^{-5} M}$ $\ce{NaOH}$ solution to a $\pu{100 mL}$ $\pu{2.5\times10^{-5} M}$ $\ce{FeCl2}$ solution, green colored $\ce{Fe(OH)2}$ precipitate is found. What is the ionic product of $\ce{Fe(OH)2}$?

My book's solution:

The concentration of $\ce{Fe^2+}$ will be the same as the concentration of $\ce{FeCl2}$. So, using the formula for dilution we can get the concentration of $\ce{Fe^2+}$:

$$\begin{align} M_1V_1 &= M_2V_2 \\ 2.5\times10^5\times100 &= M_2\times250 \\ M_2 &= \frac{100\times2.5\times10^{-5}}{250} \\ &= \pu{1\times10^{-5} M} \end{align}$$

In the same way, the concentration of $\ce{OH-}$ will be the same as the concentration of $\ce{NaOH}$. So, using the formula for dilution we can get the concentration of $\ce{OH-}$:

$$\begin{align} M_3V_3 &= M_4V_4 \\ M_4 &= \frac{150 \times 6.7 \times 10^{-5}}{250} \\ &= \pu{4.02\times10^-5 M} \end{align}$$

Now, ionic product of $\ce{Fe(OH)2}$, $K_\mathrm{ip}$,

$$K_\mathrm{ip} = [\ce{Fe^2+}][\ce{OH-}]^2 =1.616\times10^{-14} \quad \text{[Ans.]}$$

My observations:

My book assumed that $\ce{Fe(OH)2}$ remains completely dissociated and thus the concentration of $\ce{Fe^2+}$ will be the same as the concentration of $\ce{FeCl2}$, and the concentration of $\ce{OH-}$ will be the same as the concentration of $\ce{NaOH}$. My book's answer is correct only at the very beginning of the process when any $\ce{Fe(OH)2}$ precipitate hasn't formed yet. However, after the precipitate has formed, isn't my book's answer wrong?

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    $\begingroup$ For better site experience, you can find useful how-can-i-format-math-chemistry-expressions-here. ( Not to be applied to titles ). See also upright vs italic $\endgroup$
    – Poutnik
    Nov 8, 2021 at 10:29
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    $\begingroup$ The textbook uses a simplified approach with a badly worded task. It addresses the point just before precipitation start. In fact, there should be rather inequality $K_\mathrm{sp} \le \pu{1.616E−14}$, as it does not say when prcipitation started if mixing was progressive. The constant also does not count with ions trapped in precipitation. $\endgroup$
    – Poutnik
    Nov 8, 2021 at 10:34
  • $\begingroup$ @Poutnik Edited the question using \ce and \pu using the link you kindly provided. Moreover, if you post this comment as an answer, I'll accept it as my answer. $\endgroup$
    – user60158
    Nov 8, 2021 at 12:11
  • $\begingroup$ Poutnik is right. The problem should have been better written according to : When mixing small amounts of a $\ce{6.7·10^{-5} M}$ $\ce{NaOH}$ in $ 100$ mL $\ce{2.5·10^{-5}}$ $\ce{ FeCl2}$ solution, nothing happens. But when $150$ mL of this $\ce{NaOH}$ has been added, the first $\ce{Fe(OH)2}$ precipitate becomes visible. Calculate the ionic product of $\ce{Fe{OH}2}$ . $\endgroup$
    – Maurice
    Nov 8, 2021 at 12:38

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The textbook uses a simplified approach with a badly worded task.

It addresses the point just before precipitation start. In fact, there should be rather inequality $K_\mathrm{sp} \lt \pu{1.616E−14}$, as it does not say when prcipitation started if mixing was progressive.

The constant also does not count with ions trapped in precipitation.

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